# Causal inverse of $h[n]=\delta[n]-\alpha\delta[n-1]$

Find the causal inverse of $$h[n]=\delta[n]-\alpha\delta[n-1]$$

we have $$h[0]=1$$ and $$h[1]=-\alpha$$ also $$h[n]=0$$ for $$n>1$$

From the formula $$h_i[n]=\sum_{i=1}^n\frac{h[n]h_i[n-i]}{h[0]}$$

we should have the recursive difference equation $$h_i[n]=-\alpha h_i[n-1]$$ However this result is different from the book Digital Signal Processing- (Proakis)

Also notice that the book stated that $$h[n]=0$$ for $$n≥\alpha$$ which does not make sense to me

• In general it is useful to add the title and edition of a book you quote. Proakis wrote quite a few books. Nov 27, 2021 at 16:48

Of course it should be $$n\ge 2$$, that's a typo in your edition. The sign of the formula in your question is wrong. It should be
$$h[0]h_I[n]=-\sum_{k=1}^nh[k]h_I[n-k],\qquad n>0\tag{1}$$
With $$h[0]=1$$, $$h[1]=-\alpha$$, and $$h[n]=0$$ for $$n>1$$, Eq. $$(1)$$ simplifies to
$$h_I[n]=-h[1]h_I[n-1]=\alpha h_I[n-k],\qquad n>0\tag{2}$$
And since $$h_I[0]=1/h[0]=1$$ you obtain the result given in the book.
• Oh, it means that the formula on the book has also a typo. Since for $n≥1$, we have $h_I[n]*h[n]=0$ so by taking a single term of that convolution sum to the right, we get the missing negative sign? Nov 28, 2021 at 1:53
• @qcpz: that's right, by splitting up the sum and equating it to zero for $n>0$ you get the negative sign on one side of the equation. Nov 28, 2021 at 12:58